第五章VB控制结构算法总结.docx

第五章 VB控制结构算法总结1、求最大公约数P67，例5-3Private Sub Command1_Click() Dim n As Long, m As Long, r As Long n = Int(Val(Text1.Text) m = Int(Val(Text2.Text) If (m 1 Or n 1) Then Text3.Text = 请输入合法数据 Else Do r = m Mod n m = n n = r Loop While (r 0) Text3.Text = CStr(m) End IfEnd Sub2、求N！。P70 例5-4Private Sub Command1_Click() Dim N As Integer, i As Integer, Result As Long Result = 1 累乘前Result的初始值为1 N = CInt(Text1.Text) For i = 1 To N Result = Result * i Next i Label2.Caption = Str(N) + != + Str(Result)End Sub3、判断素数P72 例5-5Private Sub Command1_Click() Dim i As Integer, j As Long For i = 2 To 30 For j = 2 To Sqr(i) If i Mod j = 0 Then Exit For Next j If j Sqr(i) Then Print Str(i); End If Next iEnd Sub4、最小公倍数P78 一阅读程序题（1）5、求一个数的因子及因子和，并判断完数。P75 例5-8Private Sub Command1_Click() Dim i As Integer, j As Integer Dim sum As Integer, s As String Text1.Text = For i = 1 To 1000 sum = 0 s = i & = For j = 1 To i - 1 If i Mod j = 0 Then sum = sum + j s = s & j & + End If Next If i = sum Then Text1.Text = Text1.Text & Left(s, Len(s) - 1) & vbCrLf End If NextEnd Sub6、级数求和P75 例5-9Private Sub Command1_Click() Dim x As Double, t As Double, Eps As Double, S As Double, i As Integer i = 1 x = Val(Text1.Text) Eps = Val(Text2.Text) t = 1 S = 1 Do t = -t * x 2 / (2 * i - 1) * (2 * i) S = S + t i = i + 1 Loop Until Abs(t) = Eps Label3.Caption = cos( & x & )的值是 & SEnd Sub7、反序数P77 例5-10Private Sub Command1_Click() Dim a As Long, i As Integer Dim s As String a = InputBox(输入一个正整数, 逆向输出) s = For i = 1 To Len(CStr(a) s = Mid(CStr(a), i, 1) & , & s Next i Print s MsgBox 输出的结果为： & Left(s, Len(s) - 1), vbDefaultButton1, 逆向输出End Sub或Private Sub Command1_Click() Dim a As Long Dim s As String a = InputBox(输入一个正整数, 逆向输出) s = Do k = a Mod 10 s = s & Str(k) & , a =a10 Loop While (a 0) MsgBox 输出的结果为： & Left(s, Len(s) - 1), vbDefaultButton1, 逆向输出End Sub8、回文数。找出所有五位的回文数。Private Sub Form_Click()Dim i As Long, j As LongDim s As String, s1 As StringFor i = 10000 To 99999 s = CStr(i) s1 = For j = 1 To Len(s) s1 = Mid(s, j, 1) & s1 Next j If s = s1 Then List1.AddItem s Next iEnd Sub9、十进制转二进制Private Sub Command1_Click()Dim x As Integer, R As Integer, S As Stringx = Val(Text1)Do R = x Mod 2 x = x 2 S = CStr(R) & SLoop Until x = 0Text2 = SEnd Sub10、十进制转十六进制Private Sub Command1_Click()Dim x As Integer, R As Integer, S As Stringx = Val(Text1)Do R = x Mod 16 x = x 16 If R = 9 Then S = CStr(R) & S Else Select Case R Case 10 S = A & S Case 11 S = B & S Case 12 S = C & S Case 13 S = D & S Case 14 S = E & S Case Else S = F & S End Select End IfLoop Until x = 0Text2 = SEnd Sub11、打印菱形。Private Sub Form_Click()Dim I As Integer, J As IntegerFor I = 1 To 5 Print Tab(10 - I); For J = 1 To 2 * I - 1 Print *; Next J Print Next IFor I = 4 To 1 Step -1 Print Tab(10 - I); For J = 1 To 2 * I - 1 Print *; Next J Print Next IEnd SubP79 编程题1Private Sub Form_Click()Dim I As Integer, J As IntegerFor I = 4 To 1 Step -1 Print Tab(10 - I); For J = 1 To 2 * I - 1 Print *; Next J Print Next IFor I = 2 To 4 Print Tab(10 - I); For J = 1 To 2 * I - 1 Print *; Next J Print Next IEnd Sub12、教材：P80页 6 注意：设前后两次X的起始值为0和1，A的值由函数InputBox输入，A的平方根由函数MsgBox输出“迭代法”：在程序中用同一个变量来存放每一次推出来的值，每一次循环都执行同一组语句，给同一变量赋以新的值，即用一个新值代替旧值，这种方法称为迭代。程序中某一变量的值在不断地变化被迭代的，称为迭代变量。Private Sub Command1_Click() Dim y As Double, x As Double, x1 As Double y = Val(InputBox(请输入要求根的数, 迭代法 , 2) x = y / 2 Do x1 = x x = (x1 + y / x1) / 2 Loop Until Abs(x - x1) = 0.000001 MsgBox Str(y) + 的根是 + Str(x1) + 和- + Str(x1)End Sub13、穷举法有一数学灯谜如下： ABCD - CDC ABCA，B，C，D均为一位非负整数，找出A，B，C，D的值Private Sub Form_Click()Dim A As Integer, B As Integer, C As Integer, D As IntegerDim P As Integer, Q As Integer, R As Integer For A = 1 To 9 For B =